Problem: A person stands $10$ meters east of an intersection and watches a car driving towards the intersection from the north at $13$ meters per second. At a certain instant, the car is $24$ meters from the intersection. What is the rate of change of the distance between the car and the person at that instant (in meters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{169}{12}$ (Choice B) B $-12$ (Choice C) C $-26$ (Choice D) D $-\sqrt{269}$
Solution: Setting up the math Let... $a(t)$ denote the distance between the car and the intersection at time $t$, $b$ denote the distance between the person and the intersection (which is always $10$ meters), and $c(t)$ denote the distance between the car and the person at time $t$. $a(t)$ $b$ $c(t)$ We are given that $a'(t)=-13$ and $b=10$. Notice that $a'$ is negative since the car is getting closer to the intersection. We are also given that $a(t_0)=24$ for a specific time $t_0$. We want to find $c'(t_0)$. Relating the measures The measures relate to each other through the Pythagorean theorem: $\begin{aligned} \,[a(t)]^2+b^2&=[c(t)]^2 \\\\\\ [a(t)]^2+(10)^2&=[c(t)]^2 \end{aligned}$ We can differentiate both sides to find an expression for $c'(t)$ : $c'(t)=\dfrac{a(t)a'(t)}{c(t)}$ Using the information to solve In order to find $c'(t_0)$ we need to find $c(t_0)$. Using the Pythagorean theorem and the fact that $a(t_0)=24$ and $b=10$, we can find that $c(t_0)=26$. Now let's plug ${a(t_0)}={24}$, ${a'(t_0)}={-13}$, and ${c(t_0)}={26}$ into the expression for $c'(t_0)$ : $\begin{aligned} c'(t_0)&=\dfrac{{a(t_0)}{a'(t_0)}}{{c(t_0)}} \\\\ &=\dfrac{({24})({-13})}{({26})} \\\\ &=-12 \end{aligned}$ In conclusion, the rate of change of the distance between the car and the person at that instant is $-12$ meters per second. Since the rate of change is negative, we know that the distance is decreasing.